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The eccentric angles of the extremities of latus rectum of the ellipse a 2 x 2 ​ + b 2 y 2 ​ =1 are given by:

Question

The eccentric angles of the extremities of latus rectum of the ellipse a2x2+b2y2=1 a^2 x^2 + b^2 y^2 = 1 are given by:

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Solution

The latus rectum of an ellipse is the line segment that passes through the focus of the ellipse, is perpendicular to the major axis and both its endpoints lie on the ellipse.

The equation of the ellipse is given by a^2x^2 + b^2y^2 = 1.

The eccentricity of the ellipse is given by e = sqrt(1 - (b^2/a^2)) if a>b (ellipse is horizontal) or e = sqrt(1 - (a^2/b^2)) if b>a (ellipse is vertical).

The extremities of the latus rectum for the ellipse are given by (ae, b^2/a) and (ae, -b^2/a) for a horizontal ellipse and (a^2/b, be) and (-a^2/b, be) for a vertical ellipse.

The eccentric angles of these points are given by the inverse sine of the y-coordinate divided by b for a horizontal ellipse and the inverse sine of the x-coordinate divided by a for a vertical ellipse.

So, the eccentric angles of the extremities of the latus rectum of the ellipse are given by sin^-1(b^2/(ae)) and -sin^-1(b^2/(ae)) for a horizontal ellipse and sin^-1(a^2/(be)) and -sin^-1(a^2/(be)) for a vertical ellipse.

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